fitting line for the first part of the graph

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Mohammed Qahosh
Mohammed Qahosh 2019-10-22
评论： Adam Danz 2023-7-5
I have the following experimental data
x=[0.000421940928270042;0.000420168067226891;0.000418410041841004;0.000416666666666667;0.000414937759336100;0.000413223140495868;0.000411522633744856;0.000409836065573771;0.000408163265306122;0.000406504065040650;0.000404858299595142;0.000403225806451613;0.000401606425702811;0.000400000000000000;0.000398406374501992;0.000396825396825397;0.000395256916996047;0.000393700787401575;0.000392156862745098;0.000390625000000000;0.000389105058365759;0.000387596899224806;0.000386100386100386;0.000384615384615385;0.000383141762452107;0.000381679389312977;0.000380228136882129;0.000378787878787879;0.000377358490566038;0.000375939849624060;0.000374531835205993;0.000373134328358209;0.000371747211895911;0.000370370370370370;0.000369003690036900;0.000367647058823529;0.000366300366300366;0.000364963503649635;0.000363636363636364;0.000362318840579710;0.000361010830324910;0.000359712230215827;0.000358422939068100];
y=[-1.38899994939224;-1.35871395793660;-1.38242344659456;-1.40608139919123;-1.38422552774949;-1.37773526527217;-1.39487365118433;-1.36518337279495;-1.39978507595054;-1.38781689227391;-1.38382984300519;-1.41419979409050;-1.33826384663955;-1.35089262006801;-1.26352834469030;-1.33052407235525;-1.26786443834638;-1.23386998433884;-1.21555999623804;-1.16188856192494;-1.06920032290667;-1.01962852114161;-1.00407928991600;-0.919072639604953;-0.991469702775368;-0.815561305166114;-0.699234124812569;-0.638147300677634;-0.621360634010523;-0.573695070265892;-0.562308245326998;-0.445366441266917;-0.230576299726422;-0.220506986714244;-0.0720667385181811;-0.0396451201248960;-0.00437232408536387;0.232858731915111;0.238100819014742;0.396828802505368;0.551746643862013;0.626113326683903;0.764840522864362];
plot(x,y)
I am plotting Y(x)
The first part of this plot can be fitted using a line. Any suggestions how to do this.
Thanks in advance for your help.

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采纳的回答

Joe Vinciguerra
Joe Vinciguerra 2019-10-22
编辑：Joe Vinciguerra 2019-10-22
% let's say you want to fit the first 10 elements
ROI = (1:10);
% I'm just sorting your data by X so it plots from left to right.
% This isn't necessary but satisfies my OCD.
A = sortrows([x y],1);
x = A(:,1);
y = A(:,2);
% Extract your region of interest.
xROI = x(ROI);
yROI = y(ROI);
% perform a linear fit using a polynomial of 1 degree.
[p, S] = polyfit(xROI,yROI,1); % p are your coefficients. S is your error (if you are interested)
fitROI = polyval(p,xROI); % now take the fit and evaluate it at x values of interest
% Let's see what it looks like
figure; hold on; % create a figure and don't overwrite it
plot(x,y,'b') % plot your original data in blue
plot(xROI,yROI,'g') % plot your region of interest in green
plot(xROI,fitROI,'r') % plot to fitted line in red
  3 个评论
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Adam Danz
Adam Danz 2019-10-23
Note that this approach arbitrarily chooses what section of the line to fit. Instead, you could use a method that finds where the slope changes significantly. That's the approach in the other answer here.
Mohammed Qahosh
Mohammed Qahosh 2019-10-23
Adam Danz Yes, actually I hope two ways now. Thank you both for your help :))

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更多回答（1 个）

Adam Danz
Adam Danz 2019-10-22
You can use ischange() with the linear method to find the point where the slope changes.
This requires that the x values are in ascending order which is why we're soring them below. See comments for details.
% Sort x and y values so that x are in ascending order
[xSort, xIdx] = sort(x);
ySort = y(xIdx);
% find linear change point
chgPoint = find(ischange(ySort,'linear','MaxNumChanges',1,'SamplePoints',xSort));
% Fit line segment prior to change point
coef = polyfit(xSort(1:chgPoint),ySort(1:chgPoint),1);
% Plot results
figure();
plot(xSort,ySort, 'k-')
hold on
xline(xSort(chgPoint),'m-')
refline(coef)
xlim([min(x),max(x)])
ylim([min(y),max(y)])
legend({'data','ChangePoint','lin fit'})
title(sprintf('y = %.1fx + %.1f', coef))
191022 170111-Figure 1.png
  6 个评论
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Moses Njovana
Moses Njovana 2023-7-4
Quick one @Adam Danz. Please kindly advise if there's a way to limit the refline line plot to a certain region of the graph?
Adam Danz
Adam Danz 2023-7-5
Refline extends to the current axes limits so you could temporarily set xlim and ylim, call refline and then return the original xlim and ylim values. However, a better approach would simply be to compute the two end points at the specified bounds and plot the line using plot().

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